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Electric Field due to a Charged Cylinder

by PFStudent
Tags: charged, cylinder, electric, field
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PFStudent
#1
Sep30-07, 09:45 AM
P: 171
1. The problem statement, all variables and given/known data

Our professor gave this as sort of a challenge to find on our own; I've been stumped for a while and just can't seem to figure it out.

Find the electric field at a point [itex]{P}[/itex] a distance [itex]{Z_{0}}[/itex] along the z-axis due to a charged cylinder of radius [itex]{R_{0}}[/itex], length [itex]{L}[/itex], and uniformly distributed charge [itex]{q}[/itex]. Find the electric field for a
(a) Hollow Cylinder and for a
(b) Solid Cylinder



Yea,...my MS Paint skills could be better.

2. Relevant equations

Electric Field Equation
[tex]
{\vec{E}_{P1}} = {\frac{{k_{e}}{q_{1}}}{{\left(r_{_{1P}}\right)}^{2}}{\hat{r}_{1P}}
[/tex]

[tex]
{{E}_{P1}} = \frac{{k_{e}}{|q_{1}|}}{{\left(r_{_{1P}}\right)}^{2}}
[/tex]

[tex]
{{dE}_{P1}} = \frac{{k_{e}}{dq}}{{\left(r_{_{1P}}\right)}^{2}}
[/tex]

Charge Density Equations

[tex]
{\lambda} = \frac{q}{L}
[/tex]

[tex]
{\sigma} = \frac{q}{A}
[/tex]

[tex]
{\rho_{q}} = \frac{q}{V}
[/tex]

Electric Field due to a Charged Ring (C.R.)

[tex]
{E_{P1_{z}}} = \frac{{k_{e}}{q}{Z_{0}}}{\left({{R_{0}}^{2}}+{{Z_{0}}^{2}}\right)^{3/2}}
[/tex]

Electric Field due to a Charged Disk (C.D.)

[tex]
{E_{P1_{z}}} = \frac{{2}{k_{e}}{q}{Z_{0}}}{{{R}_{0}}^{2}}{\left[{1}-{\frac{{{Z}_{0}}}{\sqrt{\left({{R_{0}}^{2}}+{{Z_{0}}^{2}}\right)}}}\rig ht]}
[/tex]

3. The attempt at a solution

For easier notation let,

[tex]
{q} = {q_{1}}
[/tex]

(a) Hollow Cylinder

For the hollow charged cylinder, I treat the cylinder as the sum of rings of differential length [itex]{ds}[/itex] and charge [itex]{dq}[/itex]. That is, a differential segment of the ring has a length [itex]{ds}[/itex] and charge [itex]{dq}[/itex].

I recognize that,

[tex]
{{dE}_{net}} = {dE}_{P1_{z}}
[/tex]

Where,

[tex]
{dE}_{P1_{z}} = {dE}{{cos}{\beta}}
[/tex]

Now, because I am using a ring I refer to [itex]{\lambda}[/itex], so that,

[tex]
{\lambda} = {\frac{dq}{ds}}
[/tex]

However, I can already recognize that this approach is folly because I have already indicated (in the figure) [itex]{ds}[/itex] as a differential width of the cylinder as opposed to a differential length segment.

So already I am stuck, any ideas?

Particularly, I know how I want to break up the hollow cylinder (in to a sum of differential rings) however, I am not sure how to put it together. Specifically in terms of what constant: [itex]{\lambda}[/itex], [itex]{\sigma}[/itex], or [itex]{{\rho}_{q}}[/itex]? So, that I only have to integrate once to find the electric field.

Any help is appreciated.

Thanks,

-PFStudent
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Doc Al
#2
Sep30-07, 10:08 AM
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Quote Quote by PFStudent View Post
However, I can already recognize that this approach is folly because I have already indicated (in the figure) [itex]{ds}[/itex] as a differential width of the cylinder as opposed to a differential length segment.

So already I am stuck, any ideas?
I don't see the problem. Why can't [itex]dq = \lambda dz[/itex]? (Where [itex]\lambda[/itex] is the charge per unit length of the cylinder.)
PFStudent
#3
Sep30-07, 10:41 AM
P: 171
Quote Quote by Doc Al View Post
I don't see the problem. Why can't [itex]dq = \lambda dz[/itex]? (Where [itex]\lambda[/itex] is the charge per unit length of the cylinder.)
Well, because there is a problem in the sense that if I let [itex]ds[/itex] be a differential segment of the ring then that means that I have to integrate once to find the electric field contributed due to that ring. Then integrate once more using that result to find the electric field contributed by a differential hollow disk (using [itex]{\sigma}[/itex]). Then finally integrate one last time to find the electric field due to the sums of hollow disks using [itex]{{\rho}_{q}}[/itex] to find the total electric field contributed by the charged hollow cylinder.

However, if I just want to find the electric field (due to the charged cylinder) in one integral will I not have to resort to using [itex]{{\rho}_{q}}[/itex] to set up the differential piece of which I will integrate to get the whole electric field contributed by the charged cylinder in one integral?

Basically, I have to resort to using a differential hollow disk of width [itex]{ds}[/itex] and integrate that using [itex]{{\rho}_{q}}[/itex].

Or am I missing something here. To me it just does not make sense how using a differential length segment [itex]{ds}[/itex] of a differential ring will get you the electric field, I mean when you integrate that using [itex]{\lambda}[/itex] will you not just get the elctric field contributed by just that whole ring? As opposed to the whole cylinder...? Yea,...I feel like I am missing something here.

Thanks for the reply Doc Al.

Any help is appreciated.

Thanks,

-PFStudent

Doc Al
#4
Sep30-07, 11:19 AM
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Electric Field due to a Charged Cylinder

Quote Quote by PFStudent View Post
Well, because there is a problem in the sense that if I let [itex]ds[/itex] be a differential segment of the ring then that means that I have to integrate once to find the electric field contributed due to that ring.
Since you already know the field from a ring of charge, why would you redo that calculation?
PFStudent
#5
Sep30-07, 11:33 AM
P: 171
Quote Quote by Doc Al View Post
Since you already know the field from a ring of charge, why would you redo that calculation?
But that is my point, after the first integral you get the electric field due to a charged ring, will I not have to integrate once more to get the electric field due to a charged hollow disk (using [itex]{\sigma}[/itex]). After that result will I not still have to integrate yet again to actually get the electric field due to the whole charged cylinder (using [itex]{{\rho}_{q}}[/itex])?

What I am driving at is how do you find the electric field due to a hollow charged cylinder in one integral.

In particular to find the electric field due to a charged cylinder in one integral how should and what (shape) should I break up my integral in to, so that when I integrate (only once) I will be able to find the electric field due to the charged cylinder.

That is where I am stumped.

Thanks for the reply.

Thanks,

-PFStudent
Doc Al
#6
Sep30-07, 11:39 AM
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Quote Quote by PFStudent View Post
But that is my point, after the first integral you get the electric field due to a charged ring, will I not have to integrate once more to get the electric field due to a charged hollow disk (using [itex]{\sigma}[/itex]).
No. A "hollow disk" is a ring.

What I am driving at is how do you find the electric field due to a hollow charged cylinder in one integral.

In particular to find the electric field due to a charged cylinder in one integral how should and what (shape) should I break up my integral in to, so that when I integrate (only once) I will be able to find the electric field due to the charged cylinder.
Start with the field from a ring of charge. Imagine the hollow cylinder as a stack of such rings and integrate.
PFStudent
#7
Sep30-07, 01:08 PM
P: 171
Quote Quote by Doc Al View Post
No. A "hollow disk" is a ring.
Oh...ok that makes sense now.

Quote Quote by Doc Al View Post
Start with the field from a ring of charge. Imagine the hollow cylinder as a stack of such rings and integrate.
Ok, so just to see if I am doing this right,

[tex]
{{dE}_{net}} = {{dE}_{P1_{z}}}
[/tex]

[tex]
{{dE}_{P1_{z}}} = {{dE}_{P1}}{{cos}{\beta}}
[/tex]

[tex]
{{dE}_{P1}} = {\frac{{{k}_{e}}{dq}}{{\left({r}_{{}_{1P}}\right)}^{2}}}
[/tex]

[tex]
{cos}{\beta} \equiv \frac{adj.}{hyp.}
[/tex]

So,

[tex]
{cos}{\beta} = {\frac{s}{{\sqrt{{{s}^{2}}+{{{R}_{0}}^{2}}}}}}
[/tex]

[tex]
{{dE}_{P1_{z}}} = {{dE}_{P1}}{{cos}{\beta}}
[/tex]

[tex]
{{dE}_{P1_{z}}} = {\left(\frac{{k_{e}}{dq}}{{\left(r_{_{1P}}\right)}^{2}}\right)}{{\left( \frac{s}{{\sqrt{{{s}^{2}}+{{{R}_{0}}^{2}}}}}\right)}}
[/tex]

Also,

[tex]
{\lambda} = {\frac{dq}{ds}}
[/tex]

So,

[tex]
{dq} = {\lambda}{ds}
[/tex]

In addition,

[tex]
{r_{_{1P}}} = \sqrt{{s}^{2}+{R_{0}}^{2}}
[/tex]

Ok, I am up to here, however I just want to make sure I understand the integration correctly.

For the charged ring a differential (length-wise) segment of the ring was the [itex]{ds}[/itex].



However for this charged cylinder we are letting a differential width (thickness) of the ring be [itex]{ds}[/itex].



Will the integral still work anyway since we are using [itex]{\lambda}[/itex], but noting how [itex]{ds}[/itex] does not represent the same differential shape (in both the situations above)?

I get the feeling as if I should be using [itex]{\sigma}[/itex] (because we are dealing with width (thickness))...

So, why does this integration work anyway with [itex]{\lambda}[/itex] as opposed to [itex]{\sigma}[/itex] in one integral?

Thanks Doc Al.

Thanks,

-PFStudent
Doc Al
#8
Sep30-07, 01:20 PM
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Quote Quote by PFStudent View Post
Will the integral still work anyway since we are using [itex]{\lambda}[/itex], but noting how [itex]{ds}[/itex] does not represent the same differential shape (in both the situations above)?
Sure, why not? In both cases [itex]{ds}[/itex] just represents a differential element in one dimension.

I get the feeling as if I should be using [itex]{\sigma}[/itex] (because we are dealing with width (thickness))...
But we're just integrate along the z-axis, which you are representing by s. We're integrating over a length, not an area.

So, why does this integration work anyway with [itex]{\lambda}[/itex] as opposed to [itex]{\sigma}[/itex] in one integral?
Again, we are imagining a thin ring and expressing its charge in terms of the total charge on the cylinder. The total charge is q, so the charge per unit length is [itex]\lambda = q/L[/itex], and thus each thin ring will have a charge equal to [itex]\lambda ds[/itex].
PFStudent
#9
Sep30-07, 01:56 PM
P: 171
Quote Quote by Doc Al View Post
Sure, why not? In both cases [itex]{ds}[/itex] just represents a differential element in one dimension.
Oh ok. So the key thing here is noting that [itex]{ds}[/itex] just represents a differential element in one dimension. So for other problems like this if I realize that a particular differential element is in one dimension that indicates I can use the [itex]{\lambda}[/itex] relationship. Is that right?

Quote Quote by Doc Al View Post
But we're just integrate along the z-axis, which you are representing by s. We're integrating over a length, not an area.
Got it. So because we are integrating over a length not an area, that indicates the use of [itex]{\lambda}[/itex] not [itex]{\sigma}[/itex].

So, even though a cylinder has volume and surface area, that is not the point of focus for the integration. What matters is what we are integrating over, so in this case because we are integrating over a variable length, [itex]{s}[/itex] with a differential length element, [itex]{ds}[/itex] we then use, [itex]{\lambda}[/itex] rather than, [itex]{\sigma}[/itex]. Is that right?

Quote Quote by Doc Al View Post
Again, we are imagining a thin ring and expressing its charge in terms of the total charge on the cylinder. The total charge is q, so the charge per unit length is [itex]\lambda = q/L[/itex], and thus each thin ring will have a charge equal to [itex]\lambda ds[/itex].
So, although this is a uniformly charged hollow cylinder, where we can imagine the charge distribution as being evenly distributed over the surface area of the charged cylinder (because it is hollow). That does not really matter, because the point is we are integrating over a length. Is that right?

So, then because we are integrating over a length we have the charge density of the hollow charged cylinder given as,

[tex]
{\lambda} = {\frac{q}{L}}
[/tex]

Where, because the charge is uniformly distributed then [itex]{\lambda}[/itex] must be constant such that for a differential (length) element [itex]{ds}[/itex] with a differential charge [itex]{dq}[/itex]; their ratio must be the same, that is [itex]{\lambda}[/itex].

In other words,

[tex]
{\lambda} = {\frac{dq}{ds}}
[/tex]

So that,

[tex]
{\frac{dq}{ds}} = {\frac{q}{L}}
[/tex]

Is all that right?

Thanks so much for the insights Doc Al.

Thanks,

-PFStudent
Doc Al
#10
Sep30-07, 02:35 PM
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Quote Quote by PFStudent View Post
So the key thing here is noting that [itex]{ds}[/itex] just represents a differential element in one dimension. So for other problems like this if I realize that a particular differential element is in one dimension that indicates I can use the [itex]{\lambda}[/itex] relationship. Is that right?
Yep.



Got it. So because we are integrating over a length not an area, that indicates the use of [itex]{\lambda}[/itex] not [itex]{\sigma}[/itex].
Exactly.

So, even though a cylinder has volume and surface area, that is not the point of focus for the integration. What matters is what we are integrating over, so in this case because we are integrating over a variable length, [itex]{s}[/itex] with a differential length element, [itex]{ds}[/itex] we then use, [itex]{\lambda}[/itex] rather than, [itex]{\sigma}[/itex]. Is that right?
Right.



So, although this is a uniformly charged hollow cylinder, where we can imagine the charge distribution as being evenly distributed over the surface area of the charged cylinder (because it is hollow). That does not really matter, because the point is we are integrating over a length. Is that right?
Right. All we care about is how much charge is in each thin ring, resulting from our slicing up the cylinder along its length.

So, then because we are integrating over a length we have the charge density of the hollow charged cylinder given as,

[tex]
{\lambda} = {\frac{q}{L}}
[/tex]

Where, because the charge is uniformly distributed then [itex]{\lambda}[/itex] must be constant such that for a differential (length) element [itex]{ds}[/itex] with a differential charge [itex]{dq}[/itex]; their ratio must be the same, that is [itex]{\lambda}[/itex].

In other words,

[tex]
{\lambda} = {\frac{dq}{ds}}
[/tex]

So that,

[tex]
{\frac{dq}{ds}} = {\frac{q}{L}}
[/tex]

Is all that right?
Perfect!


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